Solutions to Problem Set 14

Math 101-06

10-11-2017

[Factoring]

1. Factor $x^2 - 25$ .

$$x^2 - 25 = (x + 5)(x - 5).\quad\halmos$$


2. Factor $9 x^2 - 1$ .

$$9 x^2 - 1 = (3 x - 1)(3 x + 1).\quad\halmos$$


3. Factor $9 x^2 - 25 y^2$ .

$$9 x^2 - 25 y^2 = (3 x - 5 y)(3 x + 5 y).\quad\halmos$$


4. Factor $2 x^2 + 11 x + 15$ .

$$2 x^2 + 11 x + 15 = (2 x + 5)(x + 3).\quad\halmos$$


5. Factor $3 x^2 + 7 x + 2$ .

$$3 x^2 + 7 x + 2 = (3 x + 1)(x + 2).\quad\halmos$$


6. Factor $x^3 - 8$ .

Use the formula

$$a^3 - b^3 = (a - b)(a^2 + a b + b^2).$$

In this case, $a = x$ and $b
   = 2$ . So

$$x^3 - 8 = (x - 2)(x^2 + 2 x + 4).\quad\halmos$$


7. Factor $y^3 + 1$ .

Use the formula

$$a^3 + b^3 = (a + b)(a^2 - a b + b^2).$$

In this case, $a = y$ and $b
   = 1$ . So

$$y^3 + 1 = (y + 1)(y^2 - y + 1).\quad\halmos$$


8. Factor $8 x^3 + 27$ .

Use the formula

$$a^3 + b^3 = (a + b)(a^2 - a b + b^2).$$

In this case, $a = 2 x$ and $b = 3$ . So

$$8 x^3 + 27 = (2 x + 3)(4 x^2 - 6 x + 9).\quad\halmos$$


9. Factor $a x + 3 a + b x + 3
   b$ .

Factor by grouping:

$$a x + 3 a + b x + 3 b = (a x + 3 a) + (b x + 3 b) = a(x + 3) + b(x + 3) = (a + b)(x + 3).\quad\halmos$$


10. Factor $a x + a y - b x - b
   y$ .

Factor by grouping:

$$a x + a y - b x - b y = (a x + a y) - (b x + b y) = a(x + y) - b(x + y) = (a - b)(x + y).\quad\halmos$$


11. Factor $x^3 - 4 x^2 - 9 x +
   36$ .

Factor by grouping:

$$x^3 - 4 x^2 - 9 x + 36 = (x^3 - 4 x^2) - (9 x - 36) = x^2 (x - 4) - 9 (x - 4) = (x - 4)(x^2 - 9) = (x - 4)(x + 3)(x - 3).\quad\halmos$$


I don't know that I have ever found any satisfactory answers of my own. But every time I ask it, the question is refined. ... questioning as exploration, rather than the search for certainty. - Ta-Nehisi Coates


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